Let $G$ be an Abelian group with $|G| = n$ and let $p$ be prime with $p | n$. Show that the Sylow p-subgroup of $G$ consists of $e$ and all elements whose order is a power of $p$.
Answer: By Sylow 1, $G$ contains a subgroup of order $p$, call it $P$. Thus for any $a \in P$, $\text{o}(a)|p$, thus $\text{o}(a) = 1$ or $p$.
Comment: This is where I have gotten on my own. I am not sure where to go from here or when I need to use the Abelian property of $G$.
Hint: it suffices to prove that $G$ has a unique Sylow $p$-subgroup. What happens if $G$ has two Sylow $p$-subgroups?